Question

Calculate the oxidation number of O in ${\text{OF}}_2$ and ${\text{H}}_2\text{O}$.

Answer 1

In a neutral molecule, the sum of the oxidation numbers of all the elements is equal to $0$.

The oxidation number of F (Flourine) is -1 and H (Hydrogen) is +1. [0.5 mark]

Let us assume that the oxidation number of $\text{O}$in ${\text{OF}}_2$ is $n$.
So, if we take the sum of oxidation numbers of the elements in ${\text{OF}}_2$, we get the equation:
1 $\times$(O) + 2 $\times$ (F) = $0$
$1\times n+2\times (-1)=0$
$n=+2$ [0.5 mark]

Hence, the oxidation number of oxygen in ${\text{OF}}_2$ is +2. [0.5 mark]

Let the oxidation number of O in ${\text{H}}_2\text{O}$ is $m$. [0.5 mark]
Similarly, writing the equation for ${\text{H}}_2\text{O}$ we get,
2 $\times$ (H) + 1 $\times$ (O) = $0$
$2\times (+1)+1\times m=0$
$m=-2$ [0.5 mark]

Hence, the oxidation number of oxygen in ${\text{H}}_2\text{O}$ is -2. [0.5 mark]